Question

The production department of a Vancouver City Town Council finds that it is cost efficient to replace all road signs at once. Given ? as the random variable for the lifetime of a road sign before wear and tear, assuming that ? is normally distributed with a mean of 4,000 hours and a standard deviation of 200 hours. (a) What percentage of the road signs would be damaged from between 2,800 and 2,900 hours after being replaces (b) If the department wants no more than 1% of the road signs to wear out before they are replaced, after how many hours should all the signs be replaced

Answer 1

Solution :

Given that ,

mean = = 4000

standard deviation = = 200

(a)

P(2800 < x < 2900) = P[(2800 - 4000)/ 200) < (x - ) /  < (2900 - 4000) / 200) ]

= P(-6 < z < -5.5)

= P(z < -5.5) - P(z < -6)

= 0 - 0

= 0

(b)

P(Z < -2.326) = 0.01

z = -2.326

Using z-score formula,

x = z * +

x = -2.326 * 200 + 4000 = 3535

Anwer = 3535 hours